2017 Ondo state Ss2 Joint Promotion examination Physics both (OBJ & THOERY ) answers
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2017 Ondo state Ss2 Joint Promotion examination Physics both (OBJ & THOERY ) Answers both questions and answers here
Some symbols and meaning
^ - means raise to power
/ - means over
PHYSICS OBJ:
1-10: BABBDBCDAB
11-20: DBACBCBAAD
21-30: CBACBDCBBA
31-40: BADBACACCD
COMPLETED…
PHYSICS THEORY:
PART I
Answer Any Five (5) Question From this Part.
(1a)
NS
(b)
ms-2 or m/s2
(c)
Kgm2s-2 Nm
(2)
– matter is made up of tiny particles
molecule or atom.
– matter exerts pressure.
– matter occupies space
– matter possesses mass or weight.
(3a)
Surface Tension, is defined as the ratio of tangential force, F in the surface to the length, d along which the force acts.
(b)
Surface tension can be reduced with the action of soap and detergent.
(4) T=2u sin0/g
T=2X150X0.5/10
u=150ms^-2
g=10ms^-2
Sin30^0= 0.5
=15 sec.√√
(5a)
diffusion is the process by which different forms of matter(fliuds) mix intimately with one another owing to the kinetic nature of their particle or molecules.
(5b)
– density of the substance
– mass
– pressure
– concentration
PART II
Answer three (3) Question only From this Part.
(8a)
i. Hooke’s Law states that provided the elastic limit is not exceeded, d, extension e in an elastic material e.g (wire) is proportional to the load or applied force F.
ii. Elasticity of material is one that regain its original shape and size after the disturbing force has been removed.
iii. F = Ke
Where K is the constant of the spring or wire, known as the stiffness or elastic constant.
K = F/e
(8b)
Stress = F/A ; A = πr2
r = 2 x 10 -3m.
A = 3.142 x ( 1 x 10-3)2 = 3.142 x 10-6
F = 20N
20/3.142 x 10-6 = 6365372.374 N/m2
Strain = e/i; i = 4m, e = 0.024mm
= 0.024 x 10-3m
0.024 x 10-3 = 6 x 10-9
4
W = 1/2Fe
W = 1/2Ke2
F = 6365372.374; e = 6 x 10-9
W = 6365372.374 x 6 x 10-9
F = 0.0382J
(9a) Characteristics of sound
– Pitch —> depend on frequency
– Quality —-> depend on overtone
– Loudness —-> depend on intensity
(9b)
i. Rotation of the earth about it axis
ii. Rotation of an electric fan blades
Transverse wave
(9c)
i. 2πFt = 50t (t cancel t)
F = 50/2π = 7.96HZ
iv. Period = 1/F = 1/7.96 = 0.126sec
(10a)
A machine is a device by means of which work can be done more conveniently.
(bi)
Mechanical advantage can be defined as the ratio of the load to the effort
(bii)
Velocity ratio of te machine is defines as the ratio of the distance moved by the effort to the distance moved by the load
(biii)
Efficiency is the ratio of useful work done by the machine to the total work put into the machine.
(10c)
E = work input / work output x 100%
E = Load x distance moved by load / effort x distance moved by effort = Ly/Ex
= L/E/x/y = MA/V.R
(10di)
MA/VR x 100 = 80
MA/6 x 100 = 80
MA = 6 x 80/100 = 4.8
MA = Load/effort = 300/effort = 4.8
effort = 300/4.8 = 62.5m
#Tags: Ondo State 2017 Joint Promotion Examination Physics Answer, Ondo State 2017 Joint Promotion Examination physics Question, Ondo State 2017 SS2 Govt Joint Promotion Examination Physics Answer
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